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very large time interval T w ith T  > 0 0 . T his a rea is a t most equal t o t he a rea of half the
cycle because of cancellations of the positive a nd negative areas of a sinusoid. T he second
t erm is t his a rea multiplied by 0 2 /2T w ith T  > 0 0. Clearly this term is zero, and
PI = 0
2 2 (1.5a)
2 This shows t hat a sinusoid of amplitude 0 has a power 0 /2 regardless of the value of its
frequency Wo (wo ;.!o 0) a nd p hase II. T he rms value is 0 /.;2. I f the signal frequency is zero
2
(dc or a constant signal of amplitude 0 ), t he reader can show t hat t he power is 0 . ( b) I n C hapter 4, we show t hat a s um of two sinusoids m ayor may not be periodic,
depending on whether t he r atio WI/W2 is a r ational number or not. Therefore, the period
of this signal is not known. Hence, its power will be determined by averaging its energy
over T seconds with T  > 0 0. T hus, • E xample 1 .1
Determine t he s uitable measures o f t he signals in Fig 1.2.
In Fig. 1.2a, t he signal amplitude  > 0 as It I  > 0 0. Therefore t he s uitable measure
for this signal is i ts energy E I given by 1 = 1
0 00 Ef f 2(t) d t  00 = (2)2 dt 1 1 00 + 4 e' d t =4 +4 =8 0 I n Fig. 1.2b, the signal amplitude does not  > 0 a s It I  > 0 0. However, i t is periodic, and
therefore its power exists. We c an use Eq. (1.3) t o d etermine its power. We c an simplify
t he procedure for periodic signals by observing t hat a periodic signal repeats regularly
each period (2 seconds in this case). Therefore, averaging f 2(t) over a n infinitely large
interval is identical t o averaging this q uantity over one period (2 seconds in this case).
T hus
1
P I=2 11
1 1
l(t)dt=2 11
1 2 1
t dt=3 Recall t hat t he s ignal power is t he square of its rms value. Therefore, t he rms value of
this signal is 1 /v'3. • T he first and second integrals on t he r ighthand side are t he powers of t he two sinusoids,
which are 0 1 2 /2 a nd 0 2 2 /2 as found in p art ( a). Arguing as in p art ( a), we see t hat t he
t hird t erm is zero, a nd we h avet
(1.5b)
and t he rms value is ; ,/(01 2 + 0 2 2 )/2.
We c an readily extend this result t o a s um of any number of sinusoids with distinct
frequencies. Thus, if
tThis is true only if W I # W2. I f W I = W2, the integrand of the third term contains a constant
cos (III  (12), and the third term  > 2 0 1 02 cos (81  82) as T  > 0 0. 56 1 I ntroduction t o S ignals a nd S ystems 2 0 II 12 (t) ( t) 0 t(a) 1 3 (t) 1.2 C lassification o f S ignals 57 l (t) 1 4(t) (a) t __ 0
(b) t( c) \ t 0 (d)
t(e) Quarterly G NP : The return o f recession t F ig. 1 .3 In petl:ent change; seasonally adjusted annual rates Signals for Exercise E l.l.
g
n =l where none of t he two sinusoids have identical frequencies, then
Pf = 1
2 L~ 2
..JGn (l.5c) n =l ( c) In this case the signal is complex, and we use Eq. (1.4) t o compute t he power. 1981 '82 '83 '84 '85 '86 (1.5d)
The rrns value i s is equal t o t he s um of t he powers...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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